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How To Prove ? TanA(1+Sec2A)=Tan2A

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May 21, 2018

$L H S = tan A (1 + sec 2 A)$ $LHS=tanA(1+sec2A)$

$= tan A (1 + \frac{1}{cos 2 A}) = tan A (\frac{1 + cos 2 A}{cos 2 A})$ $=tanA(1+1/(cos2A))=tanA((1+cos2A)/(cos2A))$

$= \frac{1}{cos 2 A} [tan A \cdot 2 {cos}^{2} A]$ $=1/(cos2A)[tanA*2cos^2A]$

$= \frac{1}{cos 2 A} [\frac{sin A}{cos A} \cdot 2 {cos}^{2} A]$ $=1/(cos2A)[sinA/cosA*2cos^2A]$

$= \frac{sin 2 A}{cos 2 A} = R H S$ $=(sin2A)/(cos2A)=RHS$

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